Aleks Chapter 12 | CHEM10401

The rate of a certain reaction is given by the following rate law:
rate =k[N₂]³[O₃]

Use this information to answer the questions below.

1. What is the reaction order in N₂?
2. What is the reaction order in O₃?
3. What is overall reaction order?
4. At a certain concentration of N₂ and O₃, the initial rate of reaction is 0.800 M/s. What would the initial rate of the reaction be if the concentration of N₂ were halved? Be sure your answer has the correct number of significant digits.
5. The rate of the reaction is measured to be 1.0 x 10⁴ M/s when [N₂] = 1.4 M and [O₃] = 0.36 M. Calculate the value of the rate constant. Be sure your answer has the correct number of significant digits.

Sure, I can help with that. Here are the answers to your questions:

1. The reaction order in N2 is 3. This is determined by the exponent of [N2] in the rate law.

2. The reaction order in O3 is 1. This is determined by the exponent of [O3] in the rate law.

3. The overall reaction order is 4. This is the sum of the reaction orders in N2 and O3 (3 + 1).

For the next two questions, we’ll need to use the rate law and the given information.

4. If the concentration of N2 is halved, the rate of the reaction would be reduced by a factor of 2^3 (since the reaction order in N2 is 3), because the rate is proportional to the concentration raised to the power of the reaction order. So, the new rate would be 0.800 M/s / 2^3 = 0.100 M/s.

5. To calculate the rate constant (k), we can rearrange the rate law to solve for k and substitute the given values:

k = rate/[N2]³ [O3] = (1.0×10^4 M/s) / [(1.4M)³ * 0.36M]

I hope this helps! Let me know if you have any other questions.

The rate of a certain reaction is given by the following rate law:

rate =k[NO]² [O₂]

Use this information to answer the questions below.

What is the reaction order in NO?

What is the reaction order in O2?

What is overall reaction order?

At a certain concentration of NO and O2, the initial rate of reaction is 1.0 x 103 M / s. What would the initial rate of the reaction be if the concentration of NO were halved? Be sure your answer has the correct number of significant digits.

The rate of the reaction is measured to be 0.980 M / s when [NO] = 0.74 M and [O2] = 1.3 M. Calculate the value of the rate constant. Be sure your answer has the correct number of significant digits.

Sure, I can help with that. Here are the answers to your questions:

1. If the concentration of NO is halved, the rate of the reaction would be reduced by a factor of 2² (since the reaction order in NO is 2), because the rate is proportional to the concentration raised to the power of the reaction order. So, the new rate would be

   1.0 \times 10^3 M/s / 2^2 = 2.5 \times 10^2 M/s

   .

2. To calculate the rate constant (k), we can rearrange the rate law to solve for k and substitute the given values:

k = \frac{rate}{{[NO]}^2 [O2]} = \frac{0.980 M/s}{{(0.74 M)}^2 \times 1.3 M}

I hope this helps! Let me know if you have any other questions.

The rate of a certain reaction is given by the following rate law:

rate =k[H₂] [I₂]

What is the reaction order in H2?	1
What is the reaction order in I2?	1
What is overall reaction order?
At a certain concentration of H2 and I2, the initial rate of reaction is 0.100 M / s. What would the initial rate of the reaction be if the concentration of H2 were halved? Be sure your answer has the correct number of significant digits.	Ms
The rate of the reaction is measured to be 85.0 M / s when [H2] = 0.10 M and [I2] = 0.59 M. Calculate the value of the rate constant. Be sure your answer has the correct number of significant digits.

Use this information to answer the questions below.

Sure, I can help with that. Here are the answers to your questions:

1. If the concentration of H2 is halved, the rate of the reaction would be reduced by a factor of 2 (since the reaction order in H2 is 1), because the rate is proportional to the concentration raised to the power of the reaction order. So, the new rate would be

   0.100 M/s / 2 = 0.050 M/s

   .

2. To calculate the rate constant (k), we can rearrange the rate law to solve for k and substitute the given values:

k = \frac{rate}{{[H2]} [I2]} = \frac{85.0 M/s}{{0.10 M} \times 0.59 M}

I hope this helps! Let me know if you have any other questions.

Sketch a qualitative reaction energy diagram for a chemical reaction with and without a catalyst. Assume the uncatalyzed reaction is endothermic.

Note: Because the sketches are only qualitative, the energies in them don't have to be exact. They only have to have the right relationship to each other. For example, if one energy is less than another, that fact should be clear in your sketch.

A catalyst is a substance that increases the rate of a chemical reaction, but is not consumed or produced by the reaction, and does not change the heat of reaction. You might think of a catalyst as "grease" for the "gears" of a chemical reaction: grease on the gears makes a machine go faster, but it doesn't change what the machine produces or consumes.
For example, hydrogen and oxygen react to form water like this:
H₂(g) + O₂(g) → H₂O(g)

This reaction is very exothermic. However, if you put hydrogen and oxygen gas in a box, you will see nothing happen. That's because the reaction between H₂ and O₂ is so slow at room temperature that it would take centuries to notice any change at all. Adding some powdered platinum changes this situation dramatically, however -- the reaction will speed up so much that the platinum will start to glow with the released heat. The platinum metal catalyzes -- speeds up -- the reaction between H₂ and O₂.
What does it imply about the reaction energy diagram that a catalyst speeds up a reaction? Recall that the rate of a chemical reaction is determined by its activation energy. This tells you that what a catalyst does is lower the activation energy. Since the catalyst doesn't change the heat released or absorbed by a reaction, the only difference between the two sketches you make should be that the activation energy "barrier" in the second sketch is lower. Here are the sketches:

Sketch a qualitative reaction energy diagram for a chemical reaction with and without a catalyst. Assume the uncatalyzed reaction is exothermic.

Note: Because the sketches are only qualitative, the energies in them don't have to be exact. They only have to have the right relationship to each other. For example, if one energy is less than another, that fact should be clear in your sketch.

What does it imply about the reaction energy diagram that a catalyst speeds up a reaction? Recall that the rate of a chemical reaction is determined by its activation energy. This tells you that what a catalyst does is lower the activation energy.

Since the catalyst doesn't change the heat released or absorbed by a reaction, the only difference between the two sketches you make should be that the activation energy "barrier" in the second sketch is lower.

Here are the sketches:

Identify the molecularity of each elementary reaction in the table below.

elementary reaction	molecularity
H2(g) + 2NO(g) → N2O(g) + H2O(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.
O3(g) → O2(g)+O(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.
NO3(g) + CO(g) → NO2(g) + CO2(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.

The molecularity of a reaction is determined by the number of molecules that participate as reactants in an elementary step. Here are the molecularities for each elementary reaction:

1. H2(g) + 2NO(g) → N2O(g) + H2O(g): This reaction involves three molecules (one H2 and two NO), so it is termolecular.

2. O3(g) → O2(g)+O(g): This reaction involves one molecule (O3), so it is unimolecular.

3. NO3(g) + CO(g) → NO2(g) + CO2(g): This reaction involves two molecules (one NO3 and one CO), so it is bimolecular.

I hope this helps! Let me know if you have any other questions.

Identify the molecularity of each elementary reaction in the table below.

elementary reaction	molecularity
CH3CH2SiH3(g) → C2H4(g) + SiH4(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.
H2(g) + 2I(g) → 2HI(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.
2NO2(g) → NO3(g) + NO(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.

The molecularity of a reaction is determined by the number of molecules that participate as reactants in an elementary step. Here are the molecularities for each elementary reaction:

1. CH3CH2SiH3(g) → C2H4(g) + SiH4(g): This reaction involves one molecule (CH3CH2SiH3), so it is unimolecular.

2. H2(g) + 2I(g) → 2HI(g): This reaction involves three molecules (one H2 and two I), so it is termolecular.

3. 2NO2(g) → NO3(g) + NO(g): This reaction involves two molecules (two NO2), so it is bimolecular.

I hope this helps! Let me know if you have any other questions.

Identify the molecularity of each elementary reaction in the table below.

elementary reaction	molecularity
2NO(g) + O2(g) → 2NO2(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.
CFCl3(g) → CFCl2(g) + Cl(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.
NO3(g) + CO(g) → NO2(g) + CO2(g)	unimolecular	termolecular
	bimolecular	quadrimolecular
	Cannot be determined.

The molecularity of a reaction is determined by the number of molecules that participate as reactants in an elementary step. Here are the molecularities for each elementary reaction:

1. 2NO(g) + O2(g) → 2NO2(g): This reaction involves three molecules (two NO and one O2), so it is termolecular.

2. CFCl3(g) → CFCl2(g) + Cl(g): This reaction involves one molecule (CFCl3), so it is unimolecular.

3. NO3(g) + CO(g) → NO2(g) + CO2(g): This reaction involves two molecules (one NO3 and one CO), so it is bimolecular.

I hope this helps! Let me know if you have any other questions.

Consider the following mechanism for the decomposition of ozone:
O3 (g) → O2(g) + O(g) (1)
O3(g) + O(g) → 2O2(g) (2)

1. Write the chemical equation of the overall reaction.
   

2. Are there any intermediates in this mechanism?

   • yes

   • no

3. If there are intermediates, write down their chemical formulas.
   Put a comma between each chemical formula, if there's more than one.

1. The overall reaction is obtained by adding the elementary reactions together. In this case, we get:
   2O3(g) → 3O2(g)

2. An intermediate is a species that is produced in one step of a reaction mechanism and consumed in a subsequent step. Here, O(g) is produced in the first step and consumed in the second step, so it is an intermediate.

3. So, the chemical formula of the intermediate is O.

Consider the following mechanism for the decomposition of bromine:

Br2(g) → 2Br(g) (1)
Br(g) + OCl2(g) → BrOCl(g) + Cl(g) (2)
Br(g) + Cl(g) → BrCl(g) (3)

1. Write the chemical equation of the overall reaction.

2. Are there any intermediates in this mechanism?

   • yes

   • no

3. If there are intermediates, write down their chemical formulas.
   Put a comma between each chemical formula, if there's more than one.

1. The overall reaction is obtained by adding the elementary reactions together. In this case, we get:

   Br2(g) + OCl2(g) → BrOCl(g) + BrCl(g)

2. Yes, there are intermediates in this mechanism.

3. The intermediates are species that are produced in one step of a reaction mechanism and consumed in a subsequent step. Here, Br(g) and Cl(g) are produced in the first and second steps and consumed in the second and third steps, respectively. So, the chemical formulas of the intermediates are Br and Cl.

At a certain temperature the rate of this reaction is second order in NH3 with a rate constant of 0.0662 M^-1 ·s^-1:
2NH₃(g) → N₂(g) + 3H₂(g)

Suppose a vessel contains NH₃ at a concentration of 1.23M. Calculate the concentration of NH₃ in the vessel 60.0 seconds later. You may assume no other reaction is important.

Round your answer to 2 significant digits.

Sure, here’s the calculation in a simpler format:

The rate law for a second order reaction is:

1/[A]_t = kt + 1/[A]₀

where:

• [A]_t is the concentration of NH3 at time t,

• k is the rate constant,

• [A]_0 is the initial concentration of NH3.

Given that k = 0.0662 M^-1s^-1, [A]_0 = 1.23 M, and t = 60.0 s, we can substitute these values into the equation to find [A]_t:

1 / [A]_t = (0.0662 M^-1s-1) * (60.0 s) + 1 / 1.23 M

Solving this equation will give the concentration of NH3 in the vessel 60.0 seconds later. You can use a calculator or a software like Excel or Python to solve this equation. Let me know if you need further help!

Consider a hypothetical chemical reaction:
A+B → C+D
(In this equation A, B, C and D stand for some unknown chemical formulas.)

Here is an energy diagram for the reaction:

Use the energy diagram to answer these questions.

1. What is the heat of reaction?

2. Is the reaction exothermic or endothermic?

3. Can you determine the activation energy?

4. Can you determine the activation energy of the reverse reaction? (C+D → A+B)

1. The heat of the reaction, also known as the enthalpy change (ΔH), is the difference in energy between the reactants and the products. In this case, it appears to be approximately 150 kJ/mol (250 kJ/mol for A+B and 100 kJ/mol for C+D). Since the energy level is going down, it will be -150 kJ/mol.

2. The reaction is exothermic because the energy of the products (C+D) is lower than that of the reactants (A+B). In an exothermic reaction, energy is released, which is indicated by a decrease in energy from reactants to products.

3. The activation energy is the difference in energy between the peak point on the graph and the energy of the reactants. In this case, it appears to be around 100 kJ/mol (350 kJ/mol - 250 kJ/mol).

4. To determine the activation energy of the reverse reaction (C+D → A+B), we would need to know the energy of the transition state for the reverse reaction. The activation energy for the reverse reaction (C+D → A+B) can indeed be determined from the graph. The activation energy for the reverse reaction is the difference between the energy of the transition state and the energy of the products. In this case, it appears to be around 250 kJ/mol (350 kJ/mol - 100 kJ/mol).

Carbon dioxide gas is collected at 7.0°C in an evacuated flask with a measured volume of 20.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.110atm.

Calculate the mass and number of moles of carbon dioxide gas that were collected. Be sure your answer has the correct number of significant digits.

We can use the ideal gas law, PV = nRT, to solve this problem.

Given:

• P (pressure) = 0.110 atm

• V (volume) = 20.0 L

• R (gas constant) = 0.0821 L·atm/K·mol (This is the value of R when pressure is in atm and volume is in L)

• T (temperature) = 7.0°C = 280.15 K (Converted from Celsius to Kelvin by adding 273.15)

We can rearrange the ideal gas law to solve for n (number of moles):

n = PV / RT

Substituting the given values:

n = (0.110 atm 20.0 L) / (0.0821 L·atm/K·mol 280.15 K)

Once you calculate the number of moles (n), you can find the mass of the carbon dioxide gas by multiplying the number of moles by the molar mass of CO2 (approximately 44.01 g/mol):

mass = n * molar mass of CO2

Let’s calculate the number of moles and mass of carbon dioxide gas.

First, we calculate the number of moles (n) using the ideal gas law:

n = PV / RT

Substituting the given values:

n = (0.110 atm * 20.0 L) / (0.0821 L·atm/K·mol 280.15 K)

n ≈ 0.090 mol

Next, we calculate the mass of the carbon dioxide gas by multiplying the number of moles by the molar mass of CO2 (approximately 44.01 g/mol):

mass = n * molar mass of CO2

mass = 0.090 mol * 44.01 g/mol

mass ≈ 4.0 g

So, approximately 0.090 moles and 4.0 grams of carbon dioxide gas were collected. Let me know if you have any other questions!

A reaction at 27.0°C evolves 406.mmol of carbon monoxide gas.

Calculate the volume of carbon monoxide gas that is collected. You can assume the pressure in the room is exactly 1atm. Be sure your answer has the correct number of significant digits.

We can use the ideal gas law, PV = nRT, to solve this problem.

Given:

• n (number of moles) = 406 mmol = 0.406 mol (converted from mmol to mol)

• R (gas constant) = 0.0821 L·atm/K·mol (This is the value of R when pressure is in atm and volume is in L)

• T (temperature) = 27.0°C = 300.15 K (Converted from Celsius to Kelvin by adding 273.15)

• P (pressure) = 1 atm

We can rearrange the ideal gas law to solve for V (volume):

V = nRT / P

Substituting the given values:

V = (0.406 mol * 0.0821 L·atm/K·mol 300.15 K) / 1 atm

Let’s calculate this to find the volume of carbon monoxide gas that is collected.

V = (0.406 mol 0.0821 L·atm/K·mol 300.15 K) / 1 atm

V ≈ 10.0 L

So, the volume of carbon monoxide gas that is collected is approximately 10.0 liters.

A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 11.0 cm wide and 13.2 cm high. The maximum safe pressure inside the vessel has been measured to be 8.80 MPa.

For a certain reaction the vessel may contain up to 0.105kg of dinitrogen difluoride gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Be sure your answer has the correct number of significant digits.

We can use the ideal gas law, PV = nRT, to solve this problem.

Given:

• P (pressure) = 8.80 MPa = 8.80 x 10^6 Pa (converted from MPa to Pa)

• V (volume) = πr²h = π (5.5 cm)^2 13.2 cm = 3.14 5.5^2 cm^2 13.2 cm = 1256.6 cm^3 = 0.0012566 m^3 (converted from cm^3 to m^3)

• n (number of moles) = mass / molar mass = 0.105 kg / 66.01 g/mol = 1.59 mol (converted from kg to g)

• R (gas constant) = 8.314 J/(mol·K) (This is the value of R when pressure is in Pa and volume is in m^3)

• n (number of moles) = mass / molar mass = 0.105 kg / 66.01 g/mol = 1.59 mol (converted from kg to g)

T = PV / nR

Substituting the given values:

T = (8.80 x 10^6 Pa * 0.0012566 m^3) / (1.59 mol * 8.314 J/(mol·K))

After calculating the temperature in Kelvin, we subtract 273.15 to convert it to degrees Celsius.

Let’s calculate this to find the maximum safe operating temperature.

561°C

The rate constant k for a certain reaction is measured at two different temperatures:

temperature	k
404.0°C	3.2E11
539.0°C	1.0E12

Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy E_a for this reaction.

Round your answer to 2 significant digits.

Some measurements of the initial rate of a certain reaction are given in the table below.

[N2]	[H2]	initial rate of reaction
1.70M	0.216M	0.287 M/s
8.44M	0.216M	7.07 M/s
1.70M	0.0778M	0.103 M/s

Use this information to write a rate law for this reaction, and calculate the value of the rate constant k.

Round your value for the rate constant to 2 significant digits. Also be sure your answer has the correct unit symbol.

Sure, here are the calculations in a step-by-step format:

1. We first observed the reaction order with respect to N₂ and H₂. The reaction is second order with respect to N₂ because when we doubled the concentration of N₂ (keeping H₂ constant), the rate of the reaction increased about four times. The reaction is first order with respect to H₂ because when we doubled the concentration of H₂ (keeping N₂ constant), the rate of the reaction decreased by about half.

2. With this information, we can write the rate law for this reaction as:

   \text{Rate} = k[N_2]^2[H_2]

3. We then used the data from the first experiment to calculate the value of the rate constant “k”. Here’s how we did it:

   • We substituted the rate, and the concentrations of N₂ and H₂ from the first experiment into the rate law:

     0.287 \, M/s = k(1.70 \, M)^2(0.216 \, M)

   • We then solved this equation for k:

     k = 0.287 / ((1.70)^2 * 0.216) = 0.487 \, M^{-1}s^{-1}

4. Finally, we rounded the calculated k to two significant digits, which gives:

   k = 0.49 \, M^{-1}s^{-1}

Based on the image you provided, we can determine the rate law by observing how the rate changes with the concentration of each reactant.

From experiments 1 and 2, [H2] changes while [N2] is constant:

\frac{0.356}{0.0334} = \left( \frac{5.09}{1.56} \right)^n

This gives us ( n = 2).

From experiments 1 and 3, [N2] changes while [H2] is constant:

\frac{0.213}{0.0334} = \left( \frac{5.63}{2.23} \right)^m

This gives us ( m = 1).

So, the rate law is:

Rate = k[N_2][H_2]^2

Using experiment 1 to find k,

k = \frac{Rate}{[N_2][H_2]^2}

k = \frac{0.0334 M/s}{(5.63 M)(5.09 M)^2}

k ≈ 0.0019 M^{-3}s^{-1}

So,

• The rate law for this reaction is (Rate = k[N_2][H_2]^2).

• The value of the rate constant (k) is approximately (0.0019 M^{-3}s{-1}).

Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of 0.0079M·s^−1:

2N2O(g) + 2N2(g) + O2(g)

Suppose a 350 mL flask is charged under these conditions with 450 mmol of dinitrogen monoxide. How much is left 40s later? You may assume no other reaction is important.

Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.